Show that the Husimi Q function equals $Q(\alpha)=\frac 1 \pi\langle \alpha |\hat \rho|\alpha\rangle$ from its relation with the Wigner function

Question

The Wigner function is given by

$$W(\alpha)=\frac{1}{\pi^2}\int \text{e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho \hat D(\beta) \right) \text{d}^2\beta,$$

where $\hat D(\beta)=\text e^{\beta \hat a^\dagger-\beta^*\hat a}$ is the displacement operator and $\hat \rho$ is the density matrix of the state being analysed.

The Husimi Q function is the Weierstrass transform of the Wigner function. This means that it is integrated over a Gaussian filter

$$Q(\alpha)=\frac 2 \pi \int W(\beta) \text e^{-2|\alpha-\beta|^2}\text{d}^2 \beta.$$

The most common defintion of the Husimi Q function is given by

$$Q(\alpha)=\frac 1 \pi\langle \alpha |\hat \rho|\alpha\rangle$$

but how do you simplify the Gaussian filter such that the function can be written in this form?

Answer 1

We take the Wigner function

$$W(\alpha)=\frac{1}{\pi^2}\int \text{e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho \hat D(\beta) \right) \text{d}^2\beta,$$

and write the displacement operator as $\hat D(\beta)=e^{\beta\hat a^\dagger-\beta^*\hat a}=e^{-\beta^*\hat a}e^{\beta\hat a^\dagger}e^{\frac 1 2|\beta|^2}$ using the BCH formula such that

$$W(\alpha)=\frac{1}{\pi^2}\int \text{e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho e^{-\beta^*\hat a} e^{\beta\hat a^\dagger}e^{\frac 1 2|\beta|^2} \right) \text{d}^2\beta.$$

Using the cyclic property of the trace, this can be rewritten as

$$W(\alpha)=\frac{1}{\pi^2}\int \text{e}^{\alpha \beta^*-\alpha^*\beta}e^{\frac 1 2|\beta|^2}\text{Tr}\left( \hat e^{\beta\hat a^\dagger} \hat \rho e^{-\beta^*\hat a} \right) \text{d}^2\beta.$$

The trace can be evaluated as

$$\text{Tr}\left( \hat e^{\beta\hat a^\dagger} \hat \rho e^{-\beta^*\hat a} \right)=\frac 1 \pi \int \text d^2\gamma \langle \gamma |e^{\beta \hat a^\dagger}\hat\rho e^{-\beta^*\hat a}|\gamma\rangle=\frac 1 \pi \int \text d^2\gamma \langle \gamma |e^{\beta \hat \gamma^*}\hat\rho e^{-\beta^*\gamma}|\gamma\rangle=\frac 1 \pi \int \text d^2\gamma e^{\beta \gamma^*-\beta^*\gamma} \langle \gamma |\hat\rho |\gamma\rangle.$$

Therefore, the Wigner function can be expressed as

$$W(\alpha)=\frac{1}{\pi^3}\int \int \text{e}^{(\alpha-\gamma)\beta^*- (\alpha^*-\gamma^*)\beta}e^{\frac 1 2|\beta|^2} \langle \gamma |\hat\rho |\gamma\rangle \text{d}^2\beta \text d^2\gamma .$$

By completing the square we find

$$W(\alpha)=\frac{1}{\pi^3}\int \int \text{e}^{\frac 1 2 (\beta+2(\alpha-\gamma))(\beta^*- 2(\alpha^*-\gamma^*))+2|\alpha-\gamma|^2} \langle \gamma |\hat\rho |\gamma\rangle \text{d}^2\beta \text d^2\gamma .$$

which can be simplified to

$$W(\alpha)=\frac{2}{\pi^2}\int \text{e}^{2|\alpha-\gamma|^2} \langle \gamma |\hat\rho |\gamma\rangle \text d^2\gamma .$$

This shows that

$$W(\alpha)=\frac{2}{\pi}\int \text{e}^{2|\alpha-\gamma|^2} Q(\gamma) \text d^2\gamma .$$

Answer 2

@Sunyam has mapped out your homework for you, but here are the two explicit steps that should allow you to unfold it, $$ \frac{1}{\pi^2}\int {e}^{\alpha \beta^*-\alpha^*\beta}\text{Tr}\left(\hat \rho e^{ -\beta^*\hat a} e^{\beta \hat a^\dagger } e^{|\beta|^2/2}\right) \text{d}^2\beta= \\ \frac{1}{\pi^2}\int {e}^{\alpha \beta^*-\alpha^*\beta} e^{|\beta|^2/2} \text{Tr}\left( e^{\beta \hat a^\dagger }\hat \rho e^{ -\beta^*\hat a} \right) \text{d}^2\beta ,$$ while $$ \text{Tr}\left( e^{\beta \hat a^\dagger }\hat \rho e^{ -\beta^*\hat a} \right)= \frac{1}{\pi}\int \!\! d^2\gamma ~ e^{\beta \gamma^*-\beta ^*\gamma}\langle \gamma|\hat \rho | \gamma \rangle . $$

I really don't know about "intuitive", though, unless that's what you call the Gaussian filtering.

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Edit in response to comment : Maybe this and Ch 12 of W. Schleich's book could be helpful.

The double integrals are reduced by "completing the square" and performing the double Gaussian integrals. Here is an example/confirmation in a trivial special case, $\hat \rho =|0 \rangle \langle|0$. On the one hand, $$ Q(\alpha)={1\over \pi} |\langle \alpha | 0\rangle |^2= {1\over \pi} e^{-|\alpha|^2}. $$ On the other hand, $$ W(\beta)= {1\over \pi^2} \int \!\! d^2\alpha ~ e^{-|\alpha|^2/2 + \alpha^* \beta -\alpha \beta^*}= {2\over \pi} e^{-2|\beta|^2}. $$ Plugging this into the Gaussian filter expression, $$ {4\over \pi^2}\int \!\! d^2\beta ~e^{-2|\beta|^2 -2|\beta-\alpha|^2} = {4\over \pi^2}\int \!\! d^\beta ~e^{-4|\beta|^2 -2|\alpha|^2 +4(\alpha \beta^* + \alpha^* \beta)} \\ = {4\over \pi^2}\int \!\! d^\beta ~e^{-4|\beta-\alpha/2|^2 -|\alpha|^2 }={1\over \pi} e^{-|\alpha|^2}, $$ the very previous expression, since the centering of the Gaussians is immaterial to its value.

Answer 3

Start from the definition of Wigner as $$W_\rho(\beta ) = \int\frac{\mathrm d^2\gamma}{\pi^2} \exp(\beta \bar\gamma-\bar\beta \gamma) \operatorname{tr}[\rho D(\gamma)], \quad D(\gamma)\equiv \exp(\gamma a^\dagger-\bar\gamma a).$$ Thus $$\int \mathrm d^2\beta \, W_\rho(\beta) e^{-2|\alpha-\beta|^2} = \int \mathrm d^2\gamma \underbrace{\left(\int\mathrm d^2\beta\, e^{\beta\bar\gamma-\bar\beta\gamma- 2|\alpha-\beta|^2}\right)}_{\equiv I(\gamma)} \operatorname{tr}[\rho D(\gamma)]. $$ The expression in parentheses is the Fourier transform of a Gaussian, and changing integration variables as $\beta\to\beta+\alpha$ becomes $$I(\gamma) = e^{\alpha\bar\gamma-\bar\alpha\gamma}\int \mathrm d^2\beta e^{\beta\bar\gamma-\bar\beta\gamma-2|\beta|^2}\\ = e^{\alpha\bar\gamma-\bar\alpha\gamma}\int\mathrm d\beta_2 e^{2i\beta_2\gamma_1-2\beta_2^2} \int \mathrm d\beta_1 e^{-2i\beta_1\gamma_2-2\beta_1^2} = \frac\pi2 e^{\alpha\bar\gamma-\bar\alpha\gamma} e^{-\frac12|\gamma|^2}. $$ Using this in the original expression we have $$\frac2\pi\int \mathrm d^2\beta \, W_\rho(\beta) e^{-2|\alpha-\beta|^2} = \int \mathrm d^2\gamma e^{\alpha\bar\gamma-\bar\alpha\gamma} \operatorname{tr}[\rho D_{-1}(\gamma)] = \pi \operatorname{tr}[\rho T(\alpha,-1)],$$ where I used the antinormally ordered displacement $D_{-1}(\gamma) = e^{-\frac12|\gamma|^2} D(\gamma) = e^{-\bar\gamma a}e^{\gamma a^\dagger}$, and I defined $$T(\alpha,-1) = \frac1\pi\int \mathrm d^2\gamma e^{\alpha\bar\gamma-\bar\alpha\gamma} D_{-1}(\gamma) = |\alpha\rangle\!\langle\alpha|.$$ The last identity follows e.g. from Eq. 6.6 of (Kahill and Glauber 1969, PhysRev.177.1857). We can also derive it directly from the identities in Prove that integrating a displacement operator with a Gaussian gives $\int d^2\gamma e^{-|\gamma|^2/2}D(\gamma)=\pi|0⟩\!⟨0|$. In particular, observe that $$ e^{\alpha\bar\gamma-\bar\alpha\gamma}D(\gamma) = D(\alpha) D(\gamma) D(-\alpha), \\ T(\alpha,-1)=\frac1\pi D(\alpha)\left(\int \mathrm d^2\gamma e^{-\frac12|\gamma|^2} D(\gamma) \right) D(-\alpha) = |\alpha\rangle\!\langle\alpha|.$$